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Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

Chapter 5Joint Probability Distributions

Applied Statistics and Probability for

Engineers

Sixth Edition

Douglas C. Montgomery George C. Runger

Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

Chapter 5 Title and Outline2

5Joint Probability

Distributions

5-1 Two or More Random Variables

5-1.1 Joint Probability Distributions

5-1.2 Marginal Probability Distributions

5-1.3 Conditional Probability Distributions

5-1.4 Independence

5-1.5 More Than Two Random Variables

5-2 Covariance and Correlation

5-3 Common Joint Distributions

5-3.1 Multinomial Probability Distribution

5-3.2 Bivariate Normal Distribution

5-4 Linear Functions of Random Variables

5-5 General Functions of Random Variables

5-6 Moment Generating Functions

CHAPTER OUTLINE


Learning Objectives for Chapter 5

After careful study of this chapter, you should be able to do the

following:1. Use joint probability mass functions and joint probability density

functions to calculate probabilities.

2. Calculate marginal and conditional probability distributions from joint probability distributions.

3. Interpret and calculate covariances and correlations between random variables.

4. Use the multinomial distribution to determine probabilities.

5. Properties of a bivariate normal distribution and to draw contour plots for the probability density function.

6. Calculate means and variances for linear combinations of random variables, and calculate probabilities for linear combinations of normally distributed random variables.

7. Determine the distribution of a general function of a random variable.

8. Calculate moment generating functions and use them to determine moments and distributions

3Chapter 5 Learning Objectives


Joint Probability Mass Function

Sec 5-1.1 Joint Probability Distributions 4


Joint Probability Density Function


Figure 5-2 Joint probability

density function for the random

variables X and Y. Probability

that (X, Y) is in the region R is

determined by the volume of

fXY(x,y) over the region R.

The joint probability density function for the continuous

random variables X and Y, denotes as fXY(x,y), satisfies the

following properties:


Example 5-2: Server Access Time-1Let the random variable X denote the time until a computer

server connects to your machine (in milliseconds), and let Y

denote the time until the server authorizes you as a valid user (in

milliseconds). X and Y measure the wait from a common

starting point (x < y). The joint probability density function for X

and Y is


Figure 5-4 The joint probability

density function of X and Y is

nonzero over the shaded region

where x < y.

0.001 0.002 6, for 0 and 6 10x y

XYf x y ke x y k


Example 5-2: Server Access Time-2

The region with nonzero probability is shaded in

Fig. 5-4. We verify that it integrates to 1 as follows:


0.001 0.002 0.002 0.001

0 0 0 0

0.0020.001 0.003

0 0

,

0.0030.002

10.003 1

0.003

x y y x

XY

xx x

f x y dydx ke dy dx k e dy e dx

ek e dx e dx



Now calculate a probability:


Figure 5-5 Region of

integration for the probability

that X < 1000 and Y < 2000

is darkly shaded.

1000 2000

0

1000 2000

0.002 0.001

0

1000 0.002 40.001

0

1000

0.003 4 0.001

0

3 14

1000, 2000 ,

0.002

0.003

1 10.003

0.003 0.001

XY

x

y x

x

xx

x x

P X Y f x y dydx

k e dy e dx

e ek e dx

e e e dx

e ee

0.003 316.738 11.578 0.915


Marginal Probability Distributions (discrete)

The marginal probability distribution for X is found by summing the probabilities

in each column whereas the marginal probability distribution for Y is found by

summing the probabilities in each row.

Sec 5-1.2 Marginal Probability Distributions 9

X

y

Y

x

f x f xy

f y f xy

1 2 3 f (y )

1 0.01 0.02 0.25 0.28

2 0.02 0.03 0.20 0.25

3 0.02 0.10 0.05 0.17

4 0.15 0.10 0.05 0.30

f (x ) 0.20 0.25 0.55 1.00

y = Response

time(nearest

second)

x = Number of Bars of

Signal Strength

Marginal probability distributions of X and Y


Marginal Probability Density Function (continuous)

If the joint probability density function of

random variables X and Y is fXY(x,y), the

marginal probability density functions of X

and Y are:




For the random variables that

denotes times in Example 5-2,

find the probability that Y

exceeds 2000 milliseconds.

Integrate the joint PDF directly

using the picture to determine

the limits.


2000

0 2000 2000

2000 , ,

Dark region left dark region right dark region

XY XY

x

P Y f x y dy dx f x y dy dx



Alternatively, find the marginal PDF and then

integrate that to find the desired probability.


0.001 0.002

0

0.002 0.001

0

0.0010.002

0

0.0010.002

3 0.002 0.001

0.001

1

0.001

6 10 1 for 0

y

x y

Y

y

y x

yx

y

yy

y y

f y ke dx

ke e dx

eke

eke

e e y

2000

3 0.002 0.001

2000

0.002 0.0033

2000 2000

4 63

2000

6 10 1

6 100.002 0.003

6 10 0.050.002 0.003

Y

y y

y y

P Y f y dy

e e dy

e e

e e


Mean & Variance of a Marginal Distribution

E(X) and V(X) can be obtained by first calculating the marginal

probability distribution of X and then determining E(X) and V(X) by

the usual method.


2 2

2 2

X

R

X X

R

Y

R

Y Y

R

E X x f x

V X x f x

E Y y f y

V Y y f y


Mean & Variance for Example 5-1


1 2 3

1 0.01 0.02 0.25 0.28 0.28 0.28

2 0.02 0.03 0.20 0.25 0.50 1.00

3 0.02 0.10 0.05 0.17 0.51 1.53

4 0.15 0.10 0.05 0.30 1.20 4.80

f (x ) 0.20 0.25 0.55 1.00 2.49 7.61

x *f (x ) 0.20 0.50 1.65 2.35

x 2*f (x ) 0.20 1.00 4.95 6.15

y 2*f (y )

x = Number of Bars

of Signal Strength

y = Response

time(nearest

second)

f (y ) y *f (y )

E(X) = 2.35 V(X) = 6.15 – 2.352 = 6.15 – 5.52 = 0.6275

E(Y) = 2.49 V(Y) = 7.61 – 2.492 = 7.61 – 16.20 = 1.4099


Conditional Probability Density Function

Sec 5-1.3 Conditional Probability Distributions 15


Example 5-6: Conditional Probability-1From Example 5-2, determine the conditional PDF for Y given X=x.


0.001 0.002

0.0020.001

0.0020.001

0.003

0.001 0.002

0.003

0.002 0.002

0.002

0.002

0.003 for 0

,

( ) 0.003

0.002 for 0

x y

X

x

yx

x

x

x

x yXY

Y x x

X

x y

f x k e dy

eke

eke

e x

f x y kef y

f x e

e x and x y


Example 5-6: Conditional Probability-2

Now find the probability that Y exceeds 2000 given that X=1500:


1500

2000

0.002 1500 0.002

2000

0.0023

2000

43 1

2000 1500

0.002

0.0020.002

0.002 0.3680.002

Y

y

y

P Y X

f y dy

e

ee

ee e


Mean & Variance of Conditional Random Variables

• The conditional mean of Y given X = x,

denoted as E(Y|x) or μY|x is

• The conditional variance of Y given X = x,

denoted as V(Y|x) or σ2Y|x is


Y x

y

E Y x y f y

2

2 2

Y x Y x Y x Y x

y y

V Y x y f y y f y


Example 5-8: Conditional Mean And Variance

From Example 5-2 & 5-6, what is the conditional mean for

Y given that x = 1500?


0.002 1500 0.002 3 0.002

1500 1500

0.002 0.0023

15001500

0.0023 3

1500

3

1500 0.002 0.002

0.0020.002 0.002

15000.002

0.002 0.002 0.002

0.002

y y

y y

y

E Y X y e dy e y e dy

e ee y dy

ee e

e

33

33

1500

0.002 0.002 0.002

0.002 2000 20000.002

ee

ee

If the connect time is 1500 ms, then the expected time to be authorized is 2000 ms.


Example 5-9For the discrete random variables in Exercise 5-1,

what is the conditional mean of Y given X=1?


1 2 3

1 0.01 0.02 0.25 0.28

2 0.02 0.03 0.20 0.25

3 0.02 0.10 0.05 0.17

4 0.15 0.10 0.05 0.30

f (x ) 0.20 0.25 0.55 y*f(y|x=1) y2*f(y|x=1)

1 0.050 0.080 0.455 0.05 0.05

2 0.100 0.120 0.364 0.20 0.40

3 0.100 0.400 0.091 0.30 0.90

4 0.750 0.400 0.091 3.00 12.00

Sum of f(y|x) 1.000 1.000 1.000 3.55 13.35

12.6025

0.7475

f (y )y = Response

time(nearest

second)

x = Number of Bars

of Signal Strength

The mean number of attempts given one bar is 3.55 with variance of 0.7475.


Independent Random Variables

For random variables X and Y, if any one of the

following properties is true, the others are also true.

Then X and Y are independent.

Sec 5-1.4 Independence 21


Example 5-11: Independent Random Variables

• Suppose the Example 5-2 is modified such that the joint

PDF is:

• Are X and Y independent?

• Find the probability

Sec 5-1.4 Independence 22

6 0.001 0.002, 2 10 for 0 and 0.x y

XYf x y e x y

6 0.001 0.002

0

0.001

2 10

0.001 for 0

x y

X

x

f x e dy

e x

6 0.001 0.002

0

0.002

2 10

0.002 for y > 0

x y

Y

y

f y e dx

e

1 2

1000, 1000 1000 1000

1 0.318

P X Y P X P Y

e e


Joint Probability Density Function

Sec 5-1.5 More Than Two Random Variables 23

The joint probability density function for the continuous

random variables X1, X2, X3, …Xp, denoted as

satisfies the following properties: 1 2 ... 1 2, ,...,

pX X X pf x x x


Example 5-14: Component Lifetimes

In an electronic assembly, let X1, X2, X3, X4 denote

the lifetimes of 4 components in hours. The joint

PDF is:

What is the probability that the device operates

more than 1000 hours?

The joint PDF is a product of exponential PDFs.

P(X1 > 1000, X2 > 1000, X3 > 1000, X4 > 1000)

= e-1-2-1.5-3 = e-7.5 = 0.00055


1 2 3 4

1 2 3 4

0.001 0.002 0.0015 0.00312

1 2 3 4, , , 9 10 for x 0x x x x

X X X X if x x x x e


Marginal Probability Density Function



Mean & Variance of a Joint Distribution

The mean and variance of Xi can be

determined from either the marginal PDF, or

the joint PDF as follows:



Example 5-16Points that have positive probability in the

joint probability distribution of three random

variables X1 , X2 , X3 are shown in Figure.

Suppose the 10 points are equally likely

with probability 0.1 each. The range is the

non-negative integers with x1+x2+x3 = 3

List the marginal PDF of X2


P (X2 = 0) = (3,0,0) + (0,0,3) + (1,0,2) + (2,0,1) = 0.4

P (X2 = 1) = (2,1,0) + (0,1,2) + (1,1,1) = 0.3

P (X2 = 2) = (1,2,0) + (0,2,1) = 0.2

P (X2 = 3) = (0,3,0) = 0.1

321 xxxf 321 xxxf 321 xxxf 321 xxxf

321 xxxf321 xxxf 321 xxxf

321 xxxf 321 xxxf

321 xxxf

Also, E(x2) = 0(0.4) + 1(0.3) + 2(0.2) + 3(0.1) = 1


Distribution of a Subset of Random Variables



Conditional Probability Distributions

• Conditional probability distributions can be

developed for multiple random variables by

extension of the ideas used for two random

variables.

• Suppose p = 5 and we wish to find the

distribution conditional on X4 and X5.


1 2 3 4 5

1 2 3 4 5

4 5

4 5

1 2 3 4 5

1 2 3

4 5

4 5

, , , ,, ,

,

for , 0.

X X X X X

X X X X X

X X

X X

f x x x x xf x x x

f x x

f x x


Independence with Multiple Variables

The concept of independence can be extended to

multiple variables.



Example 5-18: Layer ThicknessSuppose X1,X2, and X3 represent the thickness in μm of a

substrate, an active layer and a coating layer of a chemical

product. Assume that these variables are independent and

normally distributed with parameters and specified limits as

tabled.


X 1 X 2 X 3

Mean (μ) 10,000 1,000 80

Std dev (σ) 250 20 4

Lower limit 9,200 950 75

Upper limit 10,800 1,050 85

P(in limits) 0.99863 0.98758 0.78870

0.77783P(all in limits) =

Normal

Random VariablesParameters

and specified

limits

What proportion of the product

meets all specifications?

Answer: 0.7783, 3 layer product.

Which one of the three

thicknesses has the least

probability of meeting specs?

Answer: Layer 3 has least prob.


Covariance

• Covariance is a measure of the relationship between two random variables.

• First, we need to describe the expected value of a function of two random variables. Let h(X, Y) denote the function of interest.

Sec 5-2 Covariance & Correlation 32


Example 5-19: Expected Value of a Function of Two Random Variables


For the joint probability distribution of the two random variables in

Example 5-1, calculate E [(X-μX)(Y-μY)].

The result is obtained by multiplying x - μX times y - μY, times fxy(X,Y)

for each point in the range of (X,Y). First, μX and μy were determined

previously from the marginal distributions for X and Y:

μX = 2.35 and μy = 2.49

Therefore,


Covariance Defined



Correlation (ρ = rho)



Example 5-21: Covariance & Correlation

Determine the covariance

and correlation to the figure

below.


x y f(x, y) x-μX y-μY Prod

0 0 0.2 -1.8 -1.2 0.42

1 1 0.1 -0.8 -0.2 0.01

1 2 0.1 -0.8 0.8 -0.07

2 1 0.1 0.2 -0.2 0.00

2 2 0.1 0.2 0.8 0.02

3 3 0.4 1.2 1.8 0.88

0 0.2 1.260

1 0.2 0.926

2 0.2

3 0.4

0 0.2

1 0.2

2 0.2

3 0.4

μX = 1.8

μY = 1.8

σX = 1.1662

σY = 1.1662StD

ev

correlation =

Mar

gin

al

covariance =

Mea

n

Note the strong

positive correlation.

Join

t

Figure 5-13 Discrete joint

distribution, f(x, y).


Independence Implies ρ = 0

• If X and Y are independent random

variables,

σXY = ρXY = 0

• ρXY = 0 is necessary, but not a sufficient

condition for independence.



Example 5-23: Independence Implies Zero Covariance


Let 16 for 0 2 and 0 4

Show that 0

XY

XY

f xy x y x y

E XY E X E Y

Figure 5-15 A planar

joint distribution.

4 2

2 2

0 0

24 32

0 0

4

2

0

43

0

1

16

1

16 3

1 8

16 3

1 1 64 32

6 3 6 3 9

.

32 4 80

9 3 3

XY

E XY x y dx dy

xy dy

y dy

y

E XY E X E Y

4 2

2

0 0

24 3

0 0

42

0

4 2

2

0 0

24 22

0 0

43

0

1

16

1

16 3

1 8 1 16 4

16 2 3 6 2 3

1

16

1

16 2

2 1 64 8

16 3 8 3 3

E X x ydx dy

xy dy

y

E Y xy dx dy

xy dy

y


Multinomial Probability Distribution

• Suppose a random experiment consists of a series of n trials. Assume that:1) The outcome of each trial can be classifies into one of k

classes.

2) The probability of a trial resulting in one of the k outcomes is constant, and equal to p1, p2, …, pk.

3) The trials are independent.

• The random variables X1, X2,…, Xk denote the number of outcomes in each class and have a multinomial distribution and probability mass function:

Sec 5-3.1 Multinomial Probability Distribution 39


Example 5-25: Digital Channel

Of the 20 bits received over a digital channel, 14 are of excellent

quality, 3 are good, 2 are fair, 1 is poor. The sequence received was

EEEEEEEEEEEEEEGGGFFP. Let the random variables X1 , X2 , X3,

and X4 denote the number of bits that are E, G, F , and P, respectively,

in a transmission of 20 bits. What is the probability that 12 bits are E, 6

bits are G, 2 are F, and 0 are P?


12 6 2 0

1 2 3 4

20!12,X 6,X 2,X 0 0.6 0.3 0.08 0.02 0.0358

12!6!2!0!P X

0.03582 = (FACT(20)/(FACT(12)*FACT(6)*FACT(2))) * 0.6^12*0.3^6*0.08^2

Using Excel


Multinomial Mean and Variance

The marginal distributions of the multinomial

are binomial.

If X1, X2,…, Xk have a multinomial distribution,

the marginal probability distributions of Xi is

binomial with:

E(Xi) = npi and V(Xi) = npi(1-pi)



Bivariate Normal Probability Density Function

Sec 5-3.2 Bivariate Normal Distribution 42

2

2 2

2 22

1, ; , , , ,

2 1

21where

2 1

for and .

u

XY X X Y Y

X Y

X X Y Y

X X Y Y

f x y e

x x y yu

x y

0, ,Parameter limits: 1 1

0, ,

x x

y y

The probability density function of a bivariate normal distribution is


Marginal Distributions of the Bivariate Normal Random Variables

If X and Y have a bivariate normal distribution with

joint probability density function

fXY(x,y;σX,σY,μX,μY,ρ)

the marginal probability distributions of X and Y

are normal with means μX and μY and standard

deviations σX and σY, respectively.



Conditional Distribution of Bivariate Normal Random Variables

If X and Y have a bivariate normal distribution with

joint probability density fXY(x,y;σX,σY,μX,μY,ρ), the

conditional probability distribution of Y given X = x is

normal with mean and variance as follows:


2 2 21

YY XY x

X

YY x

x


Correlation of Bivariate Normal Random Variables

If X and Y have a bivariate normal

distribution with joint probability density

function fXY(x,y;σX,σY,μX,μY,ρ), the

correlation between X and Y is ρ.



Bivariate Normal Correlation and Independence

• In general, zero correlation does not imply independence.

• But in the special case that X and Y have a bivariate normal distribution, if ρ = 0, then X and Y are independent.

If X and Y have a bivariate normal

distribution with ρ=0, X and Y are

independent.



Linear Functions of Random Variables

• A function of random variables is itself a random variable.

• A function of random variables can be formed by either linear or nonlinear relationships. We limit our discussion here to linear functions.

• Given random variables X1, X2,…,Xp and constants c1, c2, …, cp

Y= c1X1 + c2X2 + … + cpXp

is a linear combination of X1, X2,…,Xp.

Sec 5-4 Linear Functions of Random Variables 47


Mean and Variance of a Linear Function

If X1, X2,…,Xp are random variables, and Y= c1X1 + c2X2 +

… + cpXp , then



Example 5-31: Error Propagation

A semiconductor product consists of three layers.

The variances of the thickness of each layer is 25,

40 and 30 nm. What is the variance of the finished

product?

Answer:



Mean and Variance of an Average



Reproductive Property of the Normal Distribution



Example 5-32: Linear Function of Independent Normal Random variables

Let the random variables X1 and X2 denote

the length and width of a manufactured

part. Their parameters are shown in the

table. What is the probability that the

perimeter exceeds 14.5 cm?


X 1 X 2

Mean 2 5

Std Dev 0.1 0.2

Parameters of

1 2

1 2

2 22 2

1 2

Let 2 2 perimeter

2 2 2 2 2 5 14 cm

2 2 4 0.1 4 0.2 0.04 0.16 0.20

0.20 0.4472 cm

14.5 1414.5 1 1 1.1180 0.1318

.4472

Y X X

E Y E X E X

V Y V X V X

SD Y

P Y

0.1318 = 1 - NORMDIST(14.5, 14, SQRT(0.2), TRUE)

Using Excel


General Function of a Discrete Random Variable

Suppose that X is a discrete random variable with probability distribution fX(x). Let Y = h(X) define a one-to-one transformation between the values of Xand Y so that the equation y = h(x) can be solved uniquely for x in terms of y. Let this solution be x = u(y), the inverse transform function. Then the probability mass function of the random variable Y is

fY(y) = fX[u(y)]

Sec 5-5 General Functions of Random Variables 53


Example 5-34: Function of a Discrete Random Variable

Let X be a geometric random variable with probability

distribution

fX(x) = p(1-p)x-1 , x = 1, 2, …

Find the probability distribution of Y = X2.

Solution:

– Since X ≥ 0, the transformation is one-to-one.

– The inverse transform function is X = .

– fY(y) = p(1-p) -1 , y = 1, 4, 9, 16,…


y

y


General Function of a Continuous Random Variable

Suppose that X is a continuous random variable with probability distribution fX(x). Let Y = h(X) define a one-to-one transformation between the values of X and Y so that the equation y = h(x) can be solved uniquely for x in terms of y. Let this solution be x = u(y), the inverse transform function. Then the probability distribution of Y is

fY(y) = fX[u(y)]∙|J|

where J = u’(y) is called the Jacobian of the transformation and the absolute value of J is used.



Example 5-35: Function of a Continuous Random Variable

Let X be a continuous random variable with probability

distribution:

Find the probability distribution of Y = h(X) = 2X + 4


( ) for 0 48

X

xf x x

Note that has a one-to-one relationship to .

4 1 and the Jacobian is '

2 2

4 2 1 4 for 4 12.

8 2 32Y

Y X

yx u y J u y

y yf y y


Definition of Moments about the Origin

Sec 5-6 Moment Generating Functions 57

The rth moment about the origin of the

random variable X is

( ), discrete

' ( )( ) , continuous

r

r

r r

X f x X

E XX f x dx X


Definition of a Moment-Generating Function


The moment-generating function of the random variable X is

the expected value of etX and is denoted by MX (t). That is,

( ), discrete

( ) ( )( ) , continuous

tX

tX

X tX

e f x X

M t M ee f x dx X

Let X be a random variable with moment-generating

function MX (t). Then


Example 5-36 Moment-Generating Function for a Binomial Random Variable-1


Let X follows a binomial distribution, that is

Determine the moment generating function and use it to verify that the

mean and variance of the binomial random variable are μ=np and

σ2=np(1-p).

( ) (1 ) , 0,1,....,x n xn

f x p p x nx

Now the first and second order derivatives will be

0 0

( ) (1 ) ( ) (1 )n n

tx x n x t x n x

X

x x

n nM t e p p pe p

x x

The moment-generating function is

which is the binomial expansion of

[ (1 )]t npe p

' 1

'' 2

( ) [1 ( 1)] and

( ) (1 )[1 ( 1)]

t t n

x

t t t n

x

M t npe p e

M t npe p npe p e


Example 5-36 Moment-Generating Function for a Binomial Random Variable-2


If we set t = 0 in the above two equations we get

' '

1

'' '

2

( ) and

( ) (1 )

x

x

M t np

M t np p np

Now the variance is

2 ' 2 2

2

2

(1 ) ( )

(1 )

np p np np

np np

np p

2Hence, the mean is and variance is (1 ).np np p


Properties of Moment-Generating Function


1 2

1 2

1. ( ) ( )

2. ( ) ( )

If , ,..., are independent random vari

If is a random variable

ables with

moment generat

and is a co

ing functions ( ), ( ),...,

nstant, then

( )

respectiv

n

at

X a X

aX X

n

X X X

M t e M t

M t M at

X X X

M t M t M t

X a

1 2

1 2ely, and if ... then the moment

generating function of Y is

3. ( ) ( ). ( ). ... . ( )n

n

Y X X X

Y X X X

M t M t M t M t


Example 5-38 Distribution of a Sum of Poisson Random Variables


Suppose that X1 and X2 are two independent Poisson random variables

with parameters λ1 and λ2, respectively. Determine the probability

distribution of Y = X1 + X2.

Using , the moment-generating

function of Y = X1 + X2 is

( 1)( )te

XM t e

The moment-generating function of a Poisson random variable with

parameter λ is

Hence for X1 and X2,1 2

1 2

( 1) ( 1)( ) and ( )

t te e

X XM t e M t e

1 2

1 2

1 2

( 1) ( 1)

( )( 1)

( ) ( ). ( )

t t

t

Y X X

e e

e

M t M t M t

e e

e

1 2( ) ( ). ( ). ... . ( )

nY X X XM t M t M t M t


Important Terms & Concepts for Chapter 5

Bivariate distribution

Bivariate normal distribution

Conditional mean

Conditional probability density

function

Conditional probability mass

function

Conditional variance

Contour plots

Correlation

Covariance

Error propagation

General functions of random

variables

Independence

Joint probability density function

Joint probability mass function

Linear functions of random

variables

Marginal probability distribution

Multinomial distribution

Reproductive property of the

normal distribution

Chapter 5 Summary 63

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